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Home , Factorization, Nth root, Root of unity

Chapter 3: Roots of Unity

Given a positive n, a complex z is called an of unity if zn = 1. In other words, z is a root of the Xn 1. Denote by ω , or simply − n by ω if n is understood, the e2πi/n: 2π 2π ω ω = e2πi/n cos + i sin . ≡ n ≡ n n From ωn = (e2πi/n)n = e(2πi/n)n = e2πi = 1 we see that ω is an nth . The complex 1, ω, ω2, . . . , ωn−1, (3.1)

considered as points in the , are vertices of a regular ngon inscribed in the circle. For example, when n = 6, they are vertices of a hexagon, as shown in the following figure:

Because of this particular geometric arrangement of 1, ω, ω2, . . . , ωn−1, it is not hard to believe that their sum is zero:

1 + ω + ω2 + ... + ωn−1 = 0. (3.2)

We give an algebraic proof of this extremely important identity. From ωn = 1, we have

(1 ω)(1 + ω + ω2 + ... + ωn−1) = 1 ωn = 0. − − Clearly ω = 1, that is 1 ω = 0. Hence (2) follows. − Each of ω, ω2, . . . , ωn−1 is an nth root of unity. Indeed, take z = ωk with 0 k n 1. Then ≤ ≤ − zn = (ωk)n = ωkn = (ωn)k = 1k = 1.

1 Thus 1, ω, ω2, . . . , ωn−1 are exactly n distinct roots of Xn 1. Each root ωk contributes − to a linear factor X ωk of Xn 1. Hence we have the following of Xn 1: − − − n−1 Xn 1 = (X 1)(X ω)(X ω2) (X ωn−1) = (X ωk). (3.3) − − − − − − k= 0 The last is read as the product of X ωk, where k runs from 0 to n 1. − −

Exercise 3.1. Prove that n−1 1 ωk = 1. Give a geometric interpretation of this k= 1 | − | identity. Hint: Deduce from (3.3) that n−1(X ωk) = 1 + X + X2 + + Xn−1.  k= 1 −  Notice that, since ω = 1, we have ωω = ω 2 = 1 and hence ω−1 = ω. On the | | | | other hand, ωn = 1 gives ωn−1ω = 1 and hence ωn−1 = ω−1. Thus we have ωn−1 = ω. Furthermore, from ωn−2ω2 = 1 we have ωn−2 = ω−2 = (ω−1)2 = ω2. we can continue in this manner to get ωn−3 = ω3, etc. In general, we have

ωn−k = ωk (3.4)

for k = 0, 1, . . . , n 1. −

Example 3.1. Consider the case n = 5. In this case we have

2π 2π ω = e2πi/5 cos + i sin . ≡ 5 5

We are asked to find the value of cos 2π/5.

Notice that cos 2π/5 is the real part of ω, that is, (ω + ω)/2. So it is enough to find ω + ω. In the present case, (2) gives

1 + ω + ω2 + ω3 + ω4 = 0.

On the other hand, (3.4) above gives ω4 = ω and ω3 = ω2. Thus we have 1 + ω + ω2 + ω2 + ω = 0, or 1 + (ω + ω) + (ω2 + ω2) = 0. Now

(ω + ω)2 = ω2 + ω2 + 2ωω = ω2 + ω2 + 2.

(Recall that ωω = ω 2 = 1.). So | |

ω2 + ω2 = (ω + ω)2 2. − 2 Hence 1 + (ω + ω) + (ω2 + ω2) = 0 becomes (ω + ω)2 + (ω + ω) 1 = 0. We have shown − that ω+ω 2 cos 2π/5 is a root of X2 +X 1. The roots of X2 +X 1 can be obtained ≡ − − by the usual formula for solving quadratic . The result is X = ( 1 √5)/2. − ± Since cos 2π/5 is a positive number, necessarily 2 cos 2π/5 = ( 1 + √5)/2. Thus − 2π √5 1 cos = − 5 4 which is the answer we are looking for.

Example 3.2. Now we try to factorize the polynomial X5 1 into a product of real − . In the present case, identity (3.3) becomes

X5 1 = (X 1)(X ω)(X ω2)(X ω3)(X ω4). (3.5) − − − − − − This is not the factorization we are seeking for, since X ωk (1 k 4) are not real − ≤ ≤ polynomials. Using ω4 = ω and ω3 = ω2, we rewrite this identity as

X5 1 = (X 1) (X ω)(X ω) (X ω2)(X ω2) . − − { − − }{ − − } Now (X ω)(X ω) = X2 (ω + ω)X + ωω. From the previous example we see that − − − ω + ω = (√5 1)/2 and ωω = 1. So we have − √5 1 (X ω)(X ω) = X2 − X + 1 − − − 2 which is a real polynomial. Next, we have (X ω2)(X ω2) = X2 (ω2 + ω2)X + ω2ω2. − − − Here, ω2ω2 = (ωω)2 = 1 and 2 √5 1 √5 + 1 ω2 + ω2 = (ω + ω)2 2 = − 2 = . −  2  − − 2 Substituting the last expressions into (5), we have 1 √5 1 + √5 x5 1 = (x 1) x2 + − x + 1 x2 + x + 1 , − −  2   2  which is the required factorization.

Example 3.3. We would like to find the values of cos π/12 and sin π/12. As we know, it is wise to consider eiπ/12 instead. The following computation is inspired by a 1 1 1 simple observation: = . 12 3 − 4 π π π π eπi/12 = e(πi/3)−(πi/4) = eπi/3e−πi/4 = cos + i sin cos i sin 3 3 4 − 4     1 √3 √2 √2 √6 + √2 √6 √2 = + i i = + i − . 2 2   2 − 4  2 4

3 π √6 + √2 π √6 √2 Hence we have cos = and sin = − . 12 4 12 4 Alternatively, we let ω = a + bi = eπi/12. Then

√3 1 ω2 = eπi/6 = + i. 2 2 On the other hand, ω2 = (a2 b2) + 2ab i. Comparing the real and the imaginary parts of − these two expressions of ω, we obtain

1 √3 2ab = and a2 b2 = . 2 − 2 Add the last identity to a2 + b2 = ω 2 = 1, we get | | 1 + √3 2a2 = . 2 Dividing both sides by 2 and then taking the root, we arrive at

π 1 cos = a = 1 + √3. 12 2

The reader should check that this is the same answer as the previous one for cos π/12, even though they look different.

√6 + √2 1 3.2. Verify = 1 + √3 directly. 4 2

Exercise 3.3. Plot the points of the set

Z[ω] Z + Zω ≡ consisting of all those complex numbers of the form a+bω, where a, b are , for each of the following values of ω: (a) ω = i; (b) ω = eπi/6; (c) ω = eπi/3.

Exercise 3.4. Let ω = e2πi/3 and consider the R = Z[ω] Z+Zω; for the notation, ≡ see the last exercise). Prove: (a) for each Z R, z 2 is an integer; and (b)* if an element ∈ | | z in R is invertible (in the sense that there is an element w in R such that zw = 1), then z is one of the following: 1, i, (1 + ω). (The “*” means that it is optional.) ± ± ±

Take any positive integer and let ω = e2πi/n. Consider the set

C = 1, ω, ω2, ω3, . . . , ωn−1 n { } 4 of all nth roots of unity. This set forms a with respect to in the sense

that any product of two elements in Cn is also in Cn and any element in Cn has an inverse 7 4 6 10 7+ 3 3 in Cn. For example, when n = 7, we have ω = 1 and hence ω ω = ω = ω = ω , ω−3 = ω7ω−3 = ω4, etc.

The group Cn is a in the sense that there is an element called a gen- erator, such that its powers fill the whole group. Clearly, ω is a generator. But usually this is not the only one.

More Exercises

3.5. Write out all nth roots of unity, for the following values of n: n = 2, 3, 4, 6, 8, 12.

3.6. Let ω = eπi/3. Check: (a) 1+ω3 = 0, 1+ω2 +ω4 = 0 and 1+ω+ω2 +ω3 +ω4 +ω5 = 0; (b) ω + ω5 = 1, ω ω5 = √3 i and ω4 = ω2. − 3.7. Check that if ω = eπi/m (m is a positive integer), then, for any integer k, ωk = ωm+ k. − 3.8. Verify that if ω = e2πi/5, then a = ω + ω4 and b = ω2 + ω3 are roots of t2 + t 1. − 3.9. Prove that if n 3 and if ω = 2πi/n, then 1 + ω2 + ω4 + ω6 + + ω2n−2 = 0. ≥ 3.10 Factorize the polynomials X6 1 and X8 1 into real (irreducible) polynomials as in − − Example 3.2.

3.11 Verify that if ω = e2πi/7, then a = ω + ω2 + ω4 and b = ω3 + ω5 + ω6 are roots of t2 + t + 2.

The material of the rest of this chapter is optional.

We continue with the discussion of the cyclic group Cn. Suppose that m is a positive

integer divisible by n, say n = mk, where k is another positive integer. Then Cm is generated by e2πi/m. By the substitution m = n/k, e2πi/m becomes e2πik/n ωk; (here, ≡ we still use the symbol ω for e2πi/n). Clearly a power of ωk is a power of ω. This shows

that Cm is contained in Cn. Certainly Cm is itself a group under multiplication. So l Cm

is a subgroup of Cn. To give a simple example to illustrate this, let us take n = 6 so that ω = e2πi/6 ωpii/3. Then ω6 = 1 and C = 1, ω, ω2, ω3, ω4, ω5 . Both C and C ≡ 6 { } 2 3 are subgroups of C ; in fact, C = 1, ω3 1, 1 and C = 1, ω2, ω4 1, η, η2 , 6 2 { } ≡ { − } 3 { } ≡ { } where η = ω2 e2πi/3. ≡ To recapitulate, given positive integers m and n, when m divides n (in symbols, m n), | Cm is a subgroup of Cn. Now we claim that the converse is also true:

5 If H is a subgroup of Cn, then H = Cm for some factor m of n.

To prove this, again we write ω = e2πi/n so that C = 1, ω, ω2, . . . , ωn−1 . Assume n { } that H is nontrivial, that is, H = 1 C . Let k be the smallest positive integer such { } ≡ 1 that ωk is in H. We claim that k is divisible by n. If not, we would have n = qk + r where the remainder r is nonzero, that is 0 < r < k. Now

ωr = ωn−qk = ωnωk(−q) = (ωk)−q is in H, contradicting the assumption that k is the smallest positive integer having the k property that ω is in H. Write m = n/k. Then the group Cm of all mth roots of k unity consisting of powers of ω , which is an element in H. Thus Cm is a subset of H. Conversely, if ωℓ is an element in H, then we may recycle an argument to show that ℓ is divisible by k. Now H = Cm is more or less clear.

Exercise 3.12. Suppose that r, s are positive integers divisible by n so that Cr and Cs are subgroups of Cn. As we know, the intersection of two subgroups is also a subgroup. Thus C C is a subgroup of C and hence it is of the form C for some positive r ∩ s n m number m divisible by n. What is the relation between m and the pair r and s?

Exercise 3.13. Suppose that r, s are positive integers divisible by n so that Cr and

Cs are subgroups of Cn. Let CrCs be set of all products ab, whee a is in Cr and b is

in Cs. Show that CrCs is a subgroup of Cn and hence it is of the form Cm for some positive number m divisible by n. Determine the relation between m and the pair r and s.

The number ω ω = e2πi/n is called a generator for the cyclic group C because ≡ n n every element in Cn is a power of it. But ω is usually not the only generator. Take any positive number r relative prime to n. Then a fact in elementary (which can be proved by Euclid’s ) tells us that there are integers a and b such that ar + bn = 1. Thus we have ωar+ bn = ω, which gives ωra = ω, or (ωr)a = ω. Now a k k r a general element in Cn can be written as ω for some integer k and ω = (ω ) k, r showing that every element in Cn is a power of ω . Thus we have shown that if r is relatively prime to n, then ωr is a generator.

What happens if r is not relatively prime to n? In that case there exists an integer m > 1 which is a common factor of r and n. The element ωm = e2πin/m = e2πi/s, r m r where s = n/m, generates the subgroup Cs of Cn. Since ω is a power of ω , ω

6 m belongs to the group ω generates, namely Cs, which is a proper subgroup of Cn. This r shows that ω does not generate Cn. We conclude:

2πi/n r Fact. Given ω = e and a positive integer r, ω generators Cn if and only if r and n are relatively prime.

3 2 3 4 For examples, when n = 4, ω, ω are generators of C4; when n = 5, ω, ω , ω , ω 5 are generators of C5; when n = 6, ω, ω are generators of C6; when n = 9, 2 4 5 7 8 ω, ω , ω , ω , ω , ω are generators of C9.

Generators of Cn are also called primitive nth roots of unity. For convenience, we denote by PRn the set of all primitive nth roots of unity (this is not a standard notation). Thus RP = e2πik/n GCD(k, n) = 1 . n { | } The number of elements in PRn is denoted by φ(n), that is, φ(n) = # PRn. As a of n, φ(n) is called Euler’s φ–function. It is one of the most important functions in number theory.

The nth Ψn(x) over Q is defined by

Ψn(x) = (x ω). (3.7) ω∈ RPn −  Since # PRn = ϕ(n), the degree of Ψn(x) is ϕ(n). Cyclotomic polynomials are extremely important in number theory. They have many surprising properties and we will mention some of them below. “Cyclo” literally means “circle” and “tomy” means “cutting”. It turns out that all cyclotomic polynomials have integer coefficients, and

n x 1 = Ψd(x). (3.8) − d|n  This identity tells us how to produce Ψn(x) recursively. The first six of them are 2 Ψ1(x) = x 1, Ψ2(x) = x + 1, Ψ3(x) = x + x + 1, − (3.9) Ψ (x) = x2 + 1, Ψ (x) = x4 + x3 + x2 + 1, Ψ (x) = x2 x + 1. 4 5 6 − A deep theorem in number theory says that the cyclotomic polynomials Ψn(x) are irre- ducible over the field Q of all rational numbers. Thus identity (3.8) gives the unique factorization of xn 1 into irreducible polynomials over the field of rational numbers. − Exercise 3.14. Use (3.8) and Ψ (x) = x 1 to verify (3.9). 1 −

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